I have attempted to prove: Symmetric implies diagonalisable below, in the case of real valued matrices:

$A$ is symmetric, and hence $A=A^T$. Now we know there is a jordan form for $A$, so $J=P^{-1}AP$.

Now that means $J^T = (P^{-1}AP)^T$ and thus $J^T=P^TA^T(P^{-1})^T=P^TA(P^{-1})^T$ which means that $J^T$ is a jordan block for $A$, and hence $J^T=J$(by uniqueness up to Jordan block permutation[which is irrelevant to the following argument]) and since $J$ is upper triangular, it must have no values above the diagonal, hence it is diagonal. $\blacksquare$

  • $\begingroup$ Why should $J^T$ be a Jordan canonical form matrix for $A$? $\endgroup$ – Travis Willse Jun 8 '15 at 7:33
  • $\begingroup$ Wait what was I downvoted for? Being wrong?? Isn't that the point of asking? $\endgroup$ – Eyes o.o Jun 8 '15 at 8:15
  • $\begingroup$ @user26857 fixed $\endgroup$ – Eyes o.o Jun 8 '15 at 10:07

This is the (easy) standard proof, assuming a real symmetric matrix:

Step 1) Let $v$ be an eigenvector w.r.t to the eigenvalue $a$. The computation $$a \langle v,v \rangle = \langle Av,v \rangle = \langle v,Av \rangle = \overline a \langle v,v \rangle$$ shows $a \in \mathbb R$.

Step 2) Let $U$ a non-trivial proper invariant subspace (exists by Step 1). The computation $$\langle AU^\perp,U \rangle = \langle U^\perp,AU \rangle \subset \langle U^\perp, U\rangle = 0$$ shows, that $U^\perp$ is invariant as well. Hence induction on the dimension shows the result immediately.

Step 1) is not necessary in the complex hermitian case.

  • $\begingroup$ What are the $\langle \rangle$ meaning? This is to do with an inner product? $\endgroup$ – Eyes o.o Jun 8 '15 at 8:15

In the case of a real valued matrix the existence of the Jordan form is not guaranteed, because the real numbers are not algebraically closed (i.e. the characteristic polynomial might not split), so I do not think you have provided enough reason why in this particular case the Jordan form exists.

I would suggest that you first show this is so by proving that all the characteristic values must be real, as proved in another answer here...then you can assert that $A$ does indeed have a Jordan form.

Now I can see what you are trying to say with the Jordan form: $J^T$ is also a Jordan canonical form matrix - some textbooks use the super diagonal and some the sub diagonal - and they are similar to one another. Let's say a matrix $A$ is similar to $J$ with 1's in the super diagonal, and then it would also be similar to $J^T$ - now the ones are in the sub diagonal. The "uniqueness" of the Jordan form is a tricky argument here though - $J$ is unique and $J^T$ is also unique - it does not mean they are equal though. I think if you want to go this root you should try to prove with an algebraic argument that $J=J^T$ explicitly. Then $J$ is a diagonal matrix, which is what we want, however you don't prove this ...

Lastly, the standard way to show that a symmetric matrix is similar to a diagonal matrix is to make use of Schur's theorem - it is an easier route in IMHO: if all the characteristic roots of a matrix $A$ is real (second paragraph!) then there is an orthogonal matrix $P$ so that $P^TAP$ is upper triangular...now it is easy to verify that $P^TAP$ is symmetric and therefore diagonal, and so even more can be said: $A$ is orthogonally diagonalizable.

  • 1
    $\begingroup$ The rational canonical form or as also known, classical canonical form is not the same as the Jordan form. The diagonal blocks consist of hypercompanion matrices. I am not sure whether a rational canonical matrix and its transpose being equal amounts to diagonalization, and it certainly does not seem to be what the OP had in mind. $\endgroup$ – Christiaan Hattingh Jun 8 '15 at 20:30

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